# Lesson: GMAT Counting Strategies - Part I

## Comment on GMAT Counting Strategies - Part I

### Hi Brent,

Hi Brent,

For this question - https://gmatclub.com/forum/four-women-and-three-men-must-be-seated-in-a-row-for-a-group-photo-238447.html

I am not able to understand the solution despite reading all of them especially the part "For each arrangement of 4 women, there are 5 spaces where the 3 men can be placed."

If possible, please further detail out the solution. Also, is it possible to solve using the compliment method?

Warm Regards,
Pritish

This is a super hard question. In fact, I'd say it has a 750+ level of difficulty.

Once we've arranged the 4 women (in 4! ways) we can ensure that the men are separated my only allowing each man to be seated BETWEEN two women, or on the far-left or far-right seats.

So, in the diagram _ W _ W _ W _ W _, the blanks represent the only allowable places to seat each man.
There are 5 allowable spaces where each man can be seated.

So, there are 5 allowable places to seat the first man
There are 4 allowable places to seat the second man
And there are 3 allowable places to seat the second man.

At this point, we have 2 remaining locations (aka chairs) that are unfilled.
For example, we may have: MWMW_WMW_
So, we just throw those unfilled chairs away to get: MWMWWMW

ASIDE: Solving this question via the compliment would be a nightmare since there are many different ways to break the rule about 2 men not sitting together.

Cheers,
Brent

### Hi Brent,

Hi Brent,

For this question: https://gmatclub.com/forum/ben-needs-to-form-a-committee-of-3-from-a-group-of-8-enginee-107845.html

Not entirely sure how they can come up with C1/6 for the subtraction.

So the total option is: choose 3 in a group of 8 > C3/8 = 56. If we consider 2 inexperience engineers to be one group, then there will be in total 7 groups (6 experienced engineers plus the 2 inexperienced?!) - I don't know how we can get C1/6.

Can you help? Thank you!

Here's my full solution.
Let's let A, B, C, D, E, F, G, and H represent the 8 engineers.
Let's also say A and B are the two inexperienced engineers that can't be on the committee together.
So, (# committees where A and B aren't together) = (Total # 3-person committees possible, ignoring the rule) - (# of committees where A and B are together)

Total # 3-person committees possible, ignoring the rule
Since the order in which we select the committee members does not matter, we can use combinations.
We can select 3 engineers from 8 engineers in 8C3 ways
8C3 = (8)(7)(6)/(3)(2)(1) = 56

# of committees where A and B are together
Place A and B on the committee
Select 1 engineer from the remaining 6 engineers to be on the committee with A and B.
There are 6 ways to do this (i.e, ABC, ABD, ABE, ABF, ABG, and ABH)

So, # committees where A and B aren't together = 56 - 6 = 50